To answer this question, let's go through each line of code and track the values of the variables:

• int i,j,k,l=0;: This line declares four variables i, j, k, and l. l is initialized to 0.
• k = l++;: The value of l is assigned to k first, and then l is incremented by 1. So, k becomes 0, and l becomes 1.
• j = ++k;: k is incremented by 1 first, and then the value of k is assigned to j. So, k becomes 1, and j becomes 1.
• i = j++;: The value of j is assigned to i first, and then j is incremented by 1. So, i becomes 1, and j becomes 2.
• System.out.println(i);: The value of i is printed, which is 1.

Therefore, the correct answer is B) 1.

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To answer this question, you need to understand Java modifiers.

Java modifiers are keywords that are used to specify the properties or characteristics of classes, methods, variables, and other program elements. They are used to control the visibility, accessibility, and behavior of these elements.

Let's go through each option to understand why it is correct or incorrect:

Option A) public - This option is correct. The "public" modifier is used to specify that a class, method, or variable is accessible to all other classes and code in the program.

Option B) private - This option is correct. The "private" modifier is used to specify that a class, method, or variable is only accessible within the same class.

Option C) friendly - This option is incorrect. There is no "friendly" modifier in Java. It seems like a typo or a misconception.

Option D) transient - This option is correct. The "transient" modifier is used to specify that a variable should not be serialized when an object is serialized.

The correct answer is A, B, and D. These options are correct because they are valid Java modifiers.

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To answer this question, you need to understand the concept of native methods in Java.

Option A) to access hardware that Java does not know about - This option is correct because native methods allow Java programs to access hardware-specific functionalities or interact with the underlying operating system that Java does not natively support. This is useful when you need to perform low-level operations or interact with native libraries.

Option B) to define a new data type such as an unsigned integer - This option is incorrect. Defining new data types is not the purpose of native methods. Java does not support unsigned integers by default, but you can work around this limitation using other techniques such as bit manipulation or using larger data types.

Option C) to write optimized code for performance in a language such as C/C++ - This option is correct because native methods allow you to write performance-critical code in languages like C or C++ and then integrate it into your Java program. This can be done for specific parts of the code that require high efficiency or low-level operations.

Option D) to overcome the limitation of the private scope of the method - This option is incorrect. The private scope of a method is a language-level feature in Java, and using the native keyword does not affect the scope of a method. Native methods can be declared with any access modifier, including private, but their purpose is not related to overcoming the limitation of private scope.

Therefore, the correct answers are A and C. Native methods are used to access hardware-specific functionalities and write optimized code for performance in languages like C or C++.

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To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) Compilation and output of the value 0 - This option is incorrect because the code will not compile.

Option B) Compile time error because i has not been initialized - This option is incorrect because the error in the code is not due to the initialization of variable i.

Option C) Compilation and output of null - This option is incorrect because the error in the code is not related to printing null.

Option D) Compile time error - This option is correct because there is a compile-time error in the code. The error occurs in the amethod() method where static int i; is declared. In Java, it is not allowed to declare local variables as static. Therefore, the code will fail to compile.

The correct answer is D. Compile time error.

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To answer this question, we need to understand the hierarchy of the Java Number class and its subclasses.

The Java Number class is an abstract class that serves as the superclass for all numeric wrapper classes in Java, such as Integer, Float, and Double. These wrapper classes extend the Number class and provide specific implementations for various numeric types.

Let's go through each option to determine which one does not extend java.lang.Number:

Option A) Integer - This option is incorrect because the Integer class is a wrapper class that extends the Number class. Therefore, it does extend java.lang.Number.

Option B) Boolean - This option is correct because the Boolean class does not extend the Number class. The Boolean class is a wrapper class for the primitive boolean type, which is not a numeric type.

Option C) Character - This option is correct because the Character class does not extend the Number class. The Character class is a wrapper class for the primitive char type, which is also not a numeric type.

Option D) Short - This option is incorrect because the Short class is a wrapper class that extends the Number class. Therefore, it does extend java.lang.Number.

Therefore, the correct answer is Option B (Boolean) and Option C (Character). These options do not extend java.lang.Number.

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To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) short myshort = 99S; - This option is incorrect because the suffix "S" is not a valid identifier for a short variable in Java. If you want to assign a value to a short variable, you can simply write "short myshort = 99;".

Option B) String name = 'Excellent tutorial Mr Green'; - This option is incorrect because single quotes ('') are used to represent characters in Java, not strings. To declare a string variable, you should use double quotes (""). Therefore, the correct statement would be "String name = "Excellent tutorial Mr Green";".

Option C) char c = 17c; - This option is incorrect because the suffix "c" is not a valid way to represent a character value in Java. If you want to assign a character value, you can use single quotes. For example, "char c = '17';".

Option D) int z = 015; - This option is correct because the value "015" is a valid octal (base-8) representation in Java. The leading zero indicates that the value is in octal format. Therefore, this statement will compile correctly.

The correct answer is D.

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